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[不等式] 求最小中的最大值

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lemondian posted 2024-5-18 09:28 |Read mode
给定正实数$k$,对任意正实数$a,b$,记$m=min\{a,\dfrac{b}{ka^2+b^2}\}$。求$m$的最大值。

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ic_Mivoya posted 2024-5-18 11:38
$\dfrac b{ka^2+b^2}\leqslant\dfrac b{2\sqrt{ka^2\cdot b^2}}=\dfrac1{2a\sqrt k}$

$\begin{aligned}
m&=\min\left\{a,\dfrac b{ka^2+b^2}\right\}\\
&\leqslant \min\left\{a,\dfrac1{2a\sqrt k}\right\}\\
&\leqslant\sqrt{a\cdot\dfrac1{2a\sqrt k}}\\
&=\dfrac1{\sqrt{2\sqrt k}}
\end{aligned}$

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